HDU 4405 Aeroplane chess 概率DP 水题

　　　　　　　　　　　　　　　　　　　　　Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2327    Accepted Submission(s): 1512

Problem Description

Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.

There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.

Please help Hzz calculate the expected dice throwing times to finish the game.

 

 

Input

There are multiple test cases. 
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).  
The input end with N=0, M=0. 

 

 

Output

For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.

 

 

Sample Input

2 0

8 3

2  4

4 5

7 8

0 0

 

 

Sample Output

1.1667

2.3441

 

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

 

 

 

题意：

玩一种飞行棋，有一行格子，n+1个，编号为0~n，

现在飞机从0出发，规定当飞机到达>=n的位置时，飞机到达目的地。

每次飞机要飞时，摇一个骰子，骰子数为前进的格子数，

另外格子还有很多通道，当飞机到达u时，可以直接变到v(u<v)(这是特异功能)

问飞机从0到达目的地需要给的次数（即摇骰子的次数）

 

dp[i]表示飞机从格子i到达目的地需要飞的次数。

 

 

 1 #include<cstdio>

 2 #include<cstring>

 3 

 4 using namespace std;

 5 

 6 const int MAXN=1e5+10;

 7 const int MAXM=1e3+10;

 8 const double base=1.0/6.0;

 9 

10 double dp[MAXN];

11 int jump[MAXN];

12 

13 double con(int x)

14 {

15     double res=0.0;

16     for(int i=x;i<=x+5;i++)

17         res+=dp[i];

18     return res;

19 }

20 

21 int main()

22 {

23     int n,m;

24     while(scanf("%d%d",&n,&m))

25     {

26         if(!n&&!m)

27             break;

28         memset(jump,0,sizeof(jump));

29         for(int i=0;i<m;i++)

30         {

31             int u,v;

32             scanf("%d%d",&u,&v);

33             jump[u]=v;

34         }

35         memset(dp,0,sizeof dp);

36         for(int i=n-1;i>=0;i--)

37         {

38             if(jump[i])

39                 dp[i]=dp[jump[i]];

40             else

41                 dp[i]=base*con(i+1)+1;

42         }

43         printf("%.4f\n",dp[0]);

44     }

45     return 0;

46 }

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